Explain two high school chemistry questions, 2 high school one chemistry questions, to explain and p

Question 1: 2h+ —h2

2mol 2g

mgm = the mass of hydrogen obtained after a full reaction is.

Question 2: After the mixture of magnesium, aluminum, and copper metal powders is fully reacted with excess hydrochloric acid, the products in the solution are: magnesium chloride, aluminum chloride, and copper does not react.

The filtered solution contains magnesium chloride, aluminum chloride and excess hydrochloric acid. Excessive caustic soda solution was added, and the products were magnesium hydroxide, sodium metaaluminate, water, and sodium chloride. In addition, there is an excess of sodium hydroxide in the solution.

After filtration, the ions present in the filtrate are: Na+, Cl-, AlO2-, OH-

A 1molHCl

Two Na+ AlO2-H+ Oh-Cl-mg MgCl2 mg(OH)2 precipitate.

Alal(OH)3 Naalo2 is soluble.

Cu is insoluble in acids and bases.

1 Fe + 2HCl = FeCl2 + H2 Because of the excess of iron powder, all the hydrochloric acid reacts, the ratio of HCL to H2 is 2:1, the amount of hydrochloric acid is the amount of H2 substance, and then multiplied by the molar mass, 2g mol is equal to 3g The mobile phone is too tiring, directly say the name: magnesium and hydrochloric acid react to form magnesium chloride, aluminum to produce aluminum chloride, copper does not react, add sodium hydroxide, generate magnesium hydroxide precipitation, copper hydroxide precipitation, aluminum hydroxide precipitation, after filtration, the precipitation is filtered out, because sodium hydroxide is excessive, so there must be sodium ions, hydroxide ions, chloride ions.

1. Because it is lime water, that is, Ca(OH)2 solution, Na reacts with water, the solvent decreases, and the temperature increases, so Ca(OH)2 will be precipitated, that is, it will become turbid.

However, the silvery-white color is generally the nature of the metal element, and it is impossible to replace the CA metal element.

2. Are you sure it is ferric bromide Febr3?

The ionic equation is.

2br- +cl2 = 2cl- +br2

x = 1 3, i.e. the original br-, i.e. febr3, at a concentration of 2mol l

It feels like it should be ferrous bromide Febr2, and it seems that it will be difficult.

The equation should consider the order of oxidation, the first oxidized is Fe2+, 2Fe2+ +Cl2 = 2Fe3+ +2Cl-

Then there is br-, 2br- +cl2 = br2 + 2cl-

Let Febr2 be X, then Fe2+ is X, Cl2 needs to be x 2, br- is 2x, oxidized is 2x 3, and Cl2 needs to be (2x 3) 2

The cl2 required is x 2+(2x 3) 2=

x=, i.e. the concentration is.

The total equation, i.e., febr2:cl2=:5

6Febr2 + 5Cl2-, to ensure that Fe2+ is oxidized first, and then meet the requirements of Br- to be oxidized, i.e.

6fe2+ +4br- +5cl2 = 6fe3+ +2br2 + 10cl-

Na and H2O react to produce NaOH and H2

H2O is consumed

Lime water would have been saturated.

In this way, the amount of solvent is less, and the excess solute will precipitate and become turbid.

The reason is that I don't know that ** precipitation is possible, and it is possible in all directions.

It will definitely not be silvery-white, calcium hydroxide is not silvery-white.

Fe2+ is more reducible than br-.

When Br- is oxidized, Fe2+ is completely oxidized.

CL2 is. The total number of electrons obtained is.

These electrons are given by Fe2+ and Br-.

Febr2 XMOL is available

Then the electron given by XmolFe2+ is Xmol

1 3 br- is 2x 3mol gives electrons is 2x 3mol for a total of 5x 3=

Solution x = so the concentration is.

The chemical equation is.

18Febr2 + 15Cl2 = 10FeCl3 + 8Febr3 + 6br2 ions are. 6fe(2+)+4br(-)5cl2=10cl(-)2br2+6fe(3+)

1. That's what I think.

Na reacts with water to form OH-, lime water is a Ca(OH)2 solution dissolved in water in the form of OH- and Ca2+, and there is a dissolution equilibrium.

Due to a sudden increase in OH- in solution. So Ca(OH)2That is, the solution will become cloudy. So d will appear.

And C is talking about the appearance of a silvery-white solid at the bottom of the solution. What does silvery-white mean? It's metal! In any case, there will be no metal precipitation in this reaction (Ca2+ will not be displaced, and Na will not be left).

febr3+3cl2==2fecl3+3br2

As soon as this equation came out, I thought it would be very simple, right?

CL2 knows. Isn't it possible to calculate how much febr3 is reacted to? Then calculate br-, then multiply by 3, and then divide by 3That's the answer.

If you still don't understand, welcome hi me!

Question 1 Na reacts with water to release heat, and then the temperature of the solution increases, but the solubility of Ca(OH)2 decreases, because it is saturated, so it will definitely precipitate crystals. So the solution may become cloudy.

The density of NA is smaller than that of water, and the resulting NAOH is easily soluble in water, so there is no silvery-white solid at the bottom.

2febr3 + 3cl2 = 2fecl3 + 3br2x

x = one-third of the bromine ions in the solution are oxidized to bromine element.

So the amount of ferric bromide in the solution = (

c(febr3)=

Silvery-white solid is the color of the metal sodium that has reacted with water2FeCl3 + 3Cl2 = 2FeCl3 + 3br2x

x = one-third of the bromine ions in the solution are oxidized to bromine element.

So the amount of ferric bromide in the solution = (

c(febr3)=

The first question, a, b, is a phenomenon when sodium is dissolved in water. Sodium reacts with water to reduce water, and calcium hydroxide is a micro-containment substance that may form a precipitate.

The second question, 2Febr+Cl2=BR2+2FeCl

1。The following statements about polymer compounds are incorrect (c)a. Generally has good electrical insulation.

b。Most are insoluble in water.

c.None of them are heat-resistant, and will melt and change their original shape after being heated.

d.Generally, it is stronger than metals of the same quality.

Needless to say, in C, there are two kinds of melting polymer compounds, one is thermoplastic, and the other is thermosetting. It is talked about later in Elective 5.

2。Among the following materials, which belong to the new polymer materials, are (b)aSynthetic rubber bPolymer membrane cAdhesive DPaints.

The answer seems to be uncontroversial, because the title says: "New".

1c (most polymer compounds are relatively heat-resistant, and most of them are wrong if this kind of question is too absolute).

2 A (macromolecular material, a material based on polymer compounds. Polymer materials are materials composed of compounds with high relative molecular weight, including rubber, plastics, fibers, coatings, adhesives and polymer matrix composites, and polymers are the forms of life.

All living organisms can be seen as a collection of polymers. Rubber is a class of linear flexible polymers. The secondary valence force between the molecular chains is small, the molecular chains are flexible, and the external force can produce a large deformation, and it can quickly return to its original state after removing the external force.

There are two types: natural rubber and synthetic rubber. ）

: Some proteins are soluble in water; Rubber is much less strong than metal.

2. a

1c is conserved with sodium and there is also an extreme value method. The equation for 2a is solved, and if the relative atomic mass of 2a is x, then x+

1.Select C

The amount of sodium is molar and the limit method gives the mass of the object when the product is all Na2O2, or all Na2OPick D

8n+ma=n

Subtract the two formulas to get the answer d

The first question is CSuppose that all the sodium is oxidized to sodium oxide or sodium peroxide, both are larger, so it should be a mixture of the two to be the obtained substance.

The first question is upstairs.

The second question analyzes n(al)= n(al(oh)3)=

The excess hydrochloric acid in the reaction between hydrochloric acid and elemental aluminum is.

Situation 1: If the amount of sodium hydroxide is insufficient, the consumed sodium hydroxide is divided into two parts, one part reacts with the remaining hydrochloric acid to form the other part to produce aluminum hydroxide.

Total v = case 2: if there is an excess of sodium hydroxide, the sodium hydroxide consumed is divided into three parts, one part reacts with the remaining hydrochloric acid to form the other part to form aluminum hydroxide, and the other part generates sodium metaaluminate.

Total v= This problem can also be solved with the conservation of sodium.

Situation 1: The existence of sodium is in the form of sodium chloride, and the remaining aluminum trichloride in the solution is bound chloride ions, and the chlorine bound to sodium is.

Situation 2: The amount of sodium chloride in the form of sodium chloride and sodium metaaluminate is equal to the amount of hydrochloric acid added, that is, the amount of sodium metaaluminate is.

Total. The following steps are the same as above.

Question 1:

Analysis: From the question, the mass from NaHCO3 to Na2CO3 was reduced by 9G

2NaHCO3 = Na2CO3 + CO2 + H2O difference.

xg 9gx=9*168/62=

The mass fraction of Na2CO3 is: (100% =

Question 2 Analysis: First, determine who is overdose of Al and HCL reactions.

n(al)= ；n(hcl)=c*v=

2 al +6hcl = 2alcl3 + 3h2

From the reaction equation, it can be seen that the HCl consumes mol; Generate , remaining mol

Mol HCl consumes mol NaOH ( HCl + NaOH = NaCl + H2O)1>

The generation of mol al(OH)3 consumes mol Naoh ...2>

alcl3 + 3naoh = al(oh)3 + 3nacl )

OH)3 is mol

Al(OH)3 continues to react with NaOH; Allow Al(OH)3 to dissolve.

al(oh)3 + naoh =naalo2 + 2 h2o

From the reaction equation it is seen that Al(OH)3 dissolves and consumes NaOH .3>

Obtained from <1> <2> <3> with a total consumption of NAOH

v=n/c===

1 question: 2kclo3=2kcl+3o2

2h2o2→2h2o+o2

2kmNO4 (potassium permanganate) == K2mNO4 (potassium manganate) + mnO2 (manganese dioxide) + O2 (oxygen).

Multiply the last two reactions by 3 so that the O2 coefficient is the same.

Analysis: (!) The equation transfer electrons is 6 (2) The equation is that the ox element itself is redox from -1 to 0, from -1 to -2 so the electron is 3, (3) the manganese element is reduced, only the oxygen element is increased, and the electrons transferred from the oxygen analysis are also 6 so it is 6:

The 22 questions are considered from the reaction essence ion equation.

3cu + 2no3- +8h+ =3cu2+ +4h2o +2no3 2 8 2x

According to the measurement coefficient, it is judged that the calculation of copper produces NO, (nitrate, excess hydrogen ions) v=

1.The amount of substances used to produce oxygen is the same, and the valence of oxygen in potassium permanganate and potassium chlorate is negative 2 valence, and hydrogen peroxide is negative 1 valence, and after the reaction, they are all parts, and the ratio of the three can be obtained from the conservation of electrons in redox is 2:2:1

2.It's kind of weird, but are you sure you get the title right? It may be that I can't answer the question yet.

1 C NA reacts when it comes into contact with water. will not be present at the bottom of the solution.

The solubility of D Ca(OH)2 decreases with the increase of temperature, Na reacts with water and exothermic, saturated clarified lime water has calcium hydroxide precipitation, and thus becomes turbid.

2 3 cl2+2 febr3=2 fecl3+ 3br2n(cl2)=

n(br-)=2*3*

n(febr3)=

c(febr3)=2mol/l

1. Because 2Na + 2H2O = 2NaOH + H2, (1) is the exothermic reaction, the temperature of the solution increases, and the solubility of Ca(OH)2 decreases; (2) It is solvent H2O consumption, and the solution is supersaturated. Resulting in the precipitation of a small amount of Ca(OH)2. Therefore, the phenomenon described in D will occur, and the Na reaction will be consumed and no Ca will be generated, and it is impossible for a silvery-white solid to appear at the bottom of solution C.

2. Only br- in ferric bromide (Febr3) can be oxidized. The amount of Cl2 species under standard conditions is n(Cl2)=,2br- +Cl2 = br2 + 2cl-,The amount of oxidized bromide ion species n(br-)=, it can be seen that n(br-) = 3 = in the original solution, n(febr3) = , c(febr3) = 2 mol